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\(A, B,\) and \(C\) are consecutive odd integers such that \(A < B < C.\) If \(A + B + C = 81,\) then \(A + C =\)

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\(A, B,\) and \(C\) are consecutive odd integers such that \(A < B < C.\) If \(A + B + C = 81,\) then \(A + C =\)

by Gmat_mission » Thu Sep 24, 2020 2:33 am

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\(A, B,\) and \(C\) are consecutive odd integers such that \(A < B < C.\) If \(A + B + C = 81,\) then \(A + C =\)

A) 52

B) 54

C) 56

D) 58

E) 60

Answer: B

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Re: \(A, B,\) and \(C\) are consecutive odd integers such that \(A < B < C.\) If \(A + B + C = 81,\) then \(A + C =\)

by Scott@TargetTestPrep » Thu Oct 01, 2020 9:59 am
Gmat_mission wrote: ↑
Thu Sep 24, 2020 2:33 am
 \(A, B,\) and \(C\) are consecutive odd integers such that \(A < B < C.\) If \(A + B + C = 81,\) then \(A + C =\)

A) 52

B) 54

C) 56

D) 58

E) 60

Answer: B

Solution:

Since A, B, and C are consecutive odd integers, which constitute an evenly-spaced set, we see that B must be the average, or 81/3 = 27. Thus, A = 25 and C = 29, and A + C = 25 + 29 = 54.

Answer: B

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